If you have something like [tex]\frac{1}{1+\sqrt{x}}[/tex] then you can get rid of any roots in the denominator (bottom part of the fraction) by multiplying by the conjugate [tex]1-\sqrt{x}[/tex]
Basically, the conjugate of a+b is a-b. When you multiply [tex]1+\sqrt{x}[/tex] by [tex]1-\sqrt{x}[/tex] you get [tex]1-x[/tex]. When you multiply a-b by a+b you get [tex]a^2-b^2[/tex] so you can see that if a and b are square roots, the square roots will vanish in the denominator. So multiplying by the top and the bottom will give you [tex]\frac{1}{1+\sqrt{x}}=\frac{(1-\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})}=\frac{1-\sqrt{x}}{1-x}[/tex]
That is what you call rationalizing the denominator.

Now, for your question, the conjugate of [tex]\sqrt{1-x^2}+\sqrt{1+x^2}[/tex] will be...?

If you have something like [tex]\frac{1}{1+\sqrt{x}}[/tex] then you can get rid of any roots in the denominator (bottom part of the fraction) by multiplying by the conjugate [tex]1-\sqrt{x}[/tex]
Basically, the conjugate of a+b is a-b. When you multiply [tex]1+\sqrt{x}[/tex] by [tex]1-\sqrt{x}[/tex] you get [tex]1-x[/tex]. When you multiply a-b by a+b you get [tex]a^2-b^2[/tex] so you can see that if a and b are square roots, the square roots will vanish in the denominator. So multiplying by the top and the bottom will give you [tex]\frac{1}{1+\sqrt{x}}=\frac{(1-\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})}=\frac{1-\sqrt{x}}{1-x}[/tex]
That is what you call rationalizing the denominator.

Now, for your question, the conjugate of [tex]\sqrt{1-x^2}+\sqrt{1+x^2}[/tex] will be...?

Click to expand...

Oo.. i don't know that is called conjugate. >< By rationalizing the denominator with its conjugate, now i managed to solve the question! Thanks! ^^